Ultimate Bending Capacity
for a
Reinforced Concrete Section
Under
Combined Be
nding and Axial
Load
Doug Jenkins; Interactive Design Services Pty Ltd
Strain
Stress
This paper shows the derivation of
a closed form solution for the
ultimate
bending capacity of
any
symmetrical reinforced concrete section
subject to axial load,
with any number of layers of passive
or prestressed reinforcement.
The analysis uses a rectangular compressive stress distribution in the concrete, as permitted by mo
st
design codes.
Other assumptions are
:
Plane sections remain plane.
Tensile stresses in the concrete are ignored.
Bi

l
inear elastic
/perfectly plastic
stress/strain behaviour of the reinforcement in
tension and
compression.
Specified strain at the compre
ssion face.
The analysis procedure is as follows:
1.
Select a design ultimate axial load for the section,
φ
P
u
2.
Find the axial load for “balanced stress” conditions, according to the applicable design code.
3.
Find φ, and the ultimate axial force, P
u
4.
Find the depth of the neutral axis for the force P
u
5.
Find the strain, stress, and
force in each reinforcement layer, and check that the total nett
force on the section is equal to P
u
6.
Find the total nett bending moment of the reinforcement and the concrete c
ompression
zone, about the centroid of the gross concrete section, M
u
7.
The design ultimate bending moment is then φM
u
Finding the Level of the Neutral Axis
Consider
a reinforced concrete section with:
Applied axial l
oad,
P
,
(compression positive)
and associated bending moment, M, such that
the strain at the
extreme compression face is equal to the specified ultimate compressive
strain of the concrete, ε
cu.
X axis on the section Neutral Axis, y positive towards the compression face.
Extreme compress
ion face at Y above the Neutral Axis.
Uniform concrete compressive stress
of a specified proportion of the design compressive
strength, αf’
c
, extending over a specified proportion of the region from the compressive face
to the Neutral Axis, γY
The l
evel o
f the Neutral Axis, Y, is
to be found, such that the total reaction force in the section, N, is
equal to the
nominal
applied load, P
/φ (see capacity reduction factor below)
.
For a rectangul
ar section
:
W
idth = B
Height of
zone above the Neutral Axis
= Y
Concrete stress =
αf’
c
Depth of concrete compression zone = γY
P = B.γY.
αf’
c
For a s
ection with one layer of
reinforcement:
R
einforcement
area = a
st
E
s
= Young’s Modulus of the reinforcing steel
f
ys
= Steel yield stress
ε
ps
= Steel strain due to prestr
ess
Depth below
compression
face = d
st
Steel strain = ε
cu.
(1

d
st
/Y)
Steel stress =

f
ys
<
(
ε
cu.
(1

d
st
/Y)
–
ε
ps
)
E
s
< f
ys
for layers outside the concrete compression zone or:

f
ys

αf’
c
< (ε
cu.
(1

d
st
/Y)
–
ε
ps
)E
s

αf’
c
< f
ys

αf’
c
for
layers in
side t
he concrete compression zone.
For reinforcement layers outside the concrete compression zone:
For reinforcement strain exceeding tensile yield strain:
P = B.γY.
αf’
c

f
ys
. a
st
B.γY
2
.
αf’
c

(f
ys
. a
st
+ P).Y = 0
(1.1)
For reinforcement strain
in the elastic range
:
P = B.γY.
αf’
c
+ (ε
cu.
(1

d
st
/Y)
–
ε
ps
)E
s
. a
st
B.γY
2
.
αf’
c
+ (ε
cu.
(Y

d
st
)
–
ε
ps
Y
)E
s
. a
st

PY = 0
(1.2)
For reinforcement layers inside the concrete compression zone:
For reinforcement strain
exceeding compressive yield strain:
P = B.γY.
αf’
c
+ (f
ys

αf’
c
). a
st
B.γY
2
.
αf’
c
+ ((f
ys

αf’
c
). a
st

P).Y
(1.3)
For reinforcement strain in the elastic range:
P = B.γY.
αf’
c
+
(
(ε
cu.
(1

d
st
/Y)
–
ε
ps
)E
s

αf’
c
). a
st
B.γY
2
.
αf’
c
+
(
(ε
cu.
(Y

d
st
)
–
(ε
ps
) Y)E
s

αf’
c
)
. a
st

PY = 0
(1.4)
Substituting into a
Y
2
+ bY + c
= 0
All cases:
a
=
B.γ.αf’c
(1.5
)
Equation 1.1:
b =

(f
ys
. a
st
+ P)
(1.6
)
Equation 1.2:
b =
(
ε
cu

ε
ps
)
E
s
.a
st

P
(1.7
)
Equation 1.3:
b =
((f
ys

αf’
c
). a
st

P)
(1.8
)
Equation 1.4:
b =
((
ε
cu

ε
ps
)E
s

αf’
c
).
a
st

P
(1.9)
Equations 1.1, 1.3:
c =
0
Equation
s
1.2
, 1.4
:
c =

(
ε
cu
. d
st
)E
s
.
a
st
(1.10
)
For multiple reinforcement
layers:
Sum b and c values, using equations 1.6 to 1.10
Non

rectangular concrete sections are handled by dividing the section into two layers, an upper
layer of any shape, and a rectangular or trapezoidal lower
layer
, containing the
base of the
rectangula
r stress block
.
For the top layer
:
Depth of layer = d
t
Depth of layer centroid below compression face = d
ct
Area = A
t
Then:
For a rectangular l
ayer
of width B
containing
the base of the stress block
:
P = B.(γY

d
t
).
αf’
c
Total
of the two layers:
P = (A
t
+ B.(γY

d
t
)).
αf’
c
Finally for a trapezoidal lower layer the neutral
axis equation becomes a cubic
:
aY
3
+ bY
2
+ cY
+ d
= 0
If
:
K = (B2

B1
)/D
l
B = B1
For a trapezoidal layer of top width B, containing the base of the
stress block:
P = (2B + K(γY
–
d
t
))/2 (γY

d
t
).
αf’
c
P = (K/2 . γ
2
)Y
2
+
γ(B
–
K d
t
)
Y

B
d
t
+ K
d
t
2
/2).
αf’
c
Summing the two layers, and multiplying by Y:
(
K/2 . γ
2
)Y
3
+ γ(B
–
K d
t
)
Y
2
+
(
A
t

B
d
t
+ K
d
t
2
/2
)Y
).
αf’
c
–
PY = 0
Hence the cubic
coefficients for the most general case
(including the coefficients for the
reinforcement, previously derived)
are:
All cases:
=
(
K
/
2
.
γ
2
)
α
f’
c
(2.1
)
=
γ
(
B
–
K
d
t
)
α
f’
c
(2.2
)
=
A
t
−
B
d
t
+
K
2
d
t
2
α
f’
c
−
P
st
−
P
(2.3)
Where
P
st
= Sum of reinforcement layer forces:
Equation 1.1:
𝑃
=
f
ys
.
a
st
(2.3
.1
)
Equation 1.2
:
𝑃
=
(
ε
cu
−
ε
ps
)
E
s
.
a
st
(2.3.2
)
Equation 1.3:
𝑃
=
f
ys
−
α
f’
c
)
.
a
st
(2.3.3
)
Equation 1.4:
𝑃
=
ε
cu
−
ε
ps
E
s
−
α
f’
c
)
.
a
st
(2.3.4
)
Equations 1.1, 1.3:
d =
0
Equations 1.2, 1.4:
d =

(
ε
cu
. d
st
)E
s
.
a
st
(2.4
)
Capacity Reduction Factor, φ
In the Australian codes a capacity
reduction factor, φ, is applied to the nominal axial capacity and
bending capacity to derive the design values, and the value of φ varies depending on the applied
axial load, and the balance axial load, N
ub
.
N
ub
is defined as the ultimate strength in co
mpression when k
u0
= 0.003/(0.003 + f
sy
/E
s
)
If N
u
>= N
ub
Then φ = 0.6
If N
u
< N
ub
Then φ = 0.6 + *(φ
0
–
0.6) (1
–
N
u
/ N
ub
)]
Where φ
0
is the value of φ for a section with zero axial load.
For the purposes of determining the value of φ it is normal to ass
ume that the load eccentricity, M/P
is constant. In order to find the design bending capacity, φM
u
of a section with an applied axial load
N:
Assume N = φN
u
Determine the value of N
ub
If 0 < N < N
ub
then:
φ = 0.6 + *(φ
0
–
0.6) (1
–
N
u
/ N
ub
)]
φ = 0.6 + *
(φ
0
–
0.6) (1
–
N /
φ N
ub
)]
φ
2
= 0.6
φ + *(φ
0
–
0.6) (φ
–
N / N
ub
)]
φ
2
–
(0.6
φ + *(φ
0
–
0.6) (φ
–
N / N
ub
)]) = 0
φ
2
–
φ
0
φ + (N / N
ub
)
(φ
0
–
0.6)
= 0
If (N / N
ub
)
(φ
0
–
0.6) = c
3.1
then:
φ = (φ
0
+ (φ
0
2
–
4
c)
0.5
)/2
3.2
If 4c > φ
0
2
Then N
u
> N
ub
and φ = 0.6
Example
1:
Rectangular sectio
n: 1000 mm wide x 300 mm deep, f’c
=
40 M
Pa
Reinforcement: Top, 5
No Y16, depth = 6
0 mm; Bottom,5
No Y20, depth =250 mm, E = 200 GPa
Axial load = 5
00
kN
α = 0.85
α.f’c = 0.85 * 40 = 34
γ =
0.85
–
0.007(40
–
28) = 0.766
R
einforcement areas: top = 1005 mm
2
; bottom = 1571 mm
2
k
u0
= 0.003/(0.003 + 500/200000) = 0.5455
Balance load, N
ub
= 3070 kN
c = (N / N
ub
)
(φ
0
–
0.6) = (500 / 3070)(0.8
–
0.6) = 0.03257
(3.1)
φ = (φ
0
+ (φ
0
2

4
c)
0.5
)/2
= (0.8 + (0.64
–
4*
0.03257)
0.5
) /2 =
0.7570
(3.2)
N
u
= 500000 / 0.7570 = 660525 N
a = 1000 * 0.766 * 34
=
26044
(1.5
)
Assuming that both reinforcement layers are outside the concrete compressive zone, and that
the
top reinforcement
is in the elastic rang
e
and the bottom reinforcement has yielded, apply Equation
1.1 to the bottom reinforcement, and 1.2 to the top reinforcement:
b =
(0.003
–
0)
*
200000
*1005
–
(500 * 1571)

660525
=

842746
(
1.6 and 1.7
)
c =

(
0.003 *
6
0)
*
2000
00
* 1005
=

3.619
E07
(1.10
)
Solving the quadratic
polynomial for Y yields
Y =
56.81
mm
Concrete Force = 1000 * 56.81 * 0.766 * 34 / 1000 = 1480 kN
Top reinforcement strain = 0.003 * (1

60/56.81) =

1.68E

04
Top reinfo
rcement force = 1005 * 200000 *

1.68E

04 / 1000 =

34 kN
Bottom reinforcement strain = 0.003 * (1

250/56.81) =

1.02E

02 > yield strain
Bottom reinforcement force = 1571 * 500 / 1000 =

785 kN
N
u
= 1480
–
34
–
785 = 661 kN
φN
u
= 661 * 0.7570 =
500 kN
OK
Taking moments about the centre of the section:
Concrete lever arm = 150
–
(56.81 * 0.766)/2 = 128.2 mm
Concrete moment = 1480 * 128.2 / 1000 = 190 kNm
Reinforcement moment =
(

34 * (150
–
60)
–
785 * (150
–
250)) / 1000 = 75 kNm
M
u
= 265 kNm
φM
u
= 265 * 0.7570 =
201 kNm
Example 2:
Circular section
: 600mm diameter,
f’c = 32 MPa
Reinforcement: 12 No. Y 20 bars, co
ver = 3
0 mm, E = 200 GPa
Axial load = 26
00 kN
By dividing the cross section into layers it is found that the
bottom of the concrete
compression zone
is between 300 and 350 below the compression fibre; therefore divide the circular section into two
parts:
A semi

circle of radius
300 mm.
A trapezoidal laye
r: B1 = 600 mm, B2 = 591
.6 mm, Dl =
50
mm.
α = 0.85
α.f’c = 0.85 * 32 = 27
.20 MPa
γ =
0.85
–
0.007(32
–
28) = 0.822
Reinforcement areas: top
and bottom = 314.2
mm
2
; other layers
=
628.3
mm
2
k
u0
= 0.003/(0.003 + 500/200000) = 0.5455
Balance load, N
ub
= 3021
kN
c = (N / N
ub
)
(φ
0
–
0.6)
= (2600 / 3021
)(0.8
–
0.6) = 0.
17213
(3.1)
4c > 0.64; N
u
> N
ub
; φ = 0.6
N
u
=
260
0000 / 0.
6
=
4333333
N
A
t
= П.300
2
/ 2 = 141372 mm
2
K = (591.6

600)/50 =

0.1680
The cubic coefficients are:
Solving the cub
ic polynomial for Y yields
Y = 374.7
mm
Depth of stress block = 374.7 * 0.8
22 = 308.0
Concrete Force = 3954 kN
Total reinforcement force = 380 kN
N
u
= 3954 + 380 = 4334 kN
φN
u
=
4334
* 0.
60
=
26
00 kN
OK
Taking moments about the centre of the section:
Concrete moment = 485 kNm
Reinforcement moment = 183 kNm
M
u
= 668 kNm
φM
u
=
668
* 0.
6
0 =
400
kNm
©Doug Jenkins,
23
November
2008
Layer
a
b
c
d
Concrete
1.543803
14541.9
1278313
Reo 1
148534.5
Reo 2
359901
31476364
Reo 3
359901
65973446
Reo 4
359901
113097336
Reo 5
376991
160221225
Reo 6
376991
194718307
Reo 7
188496
105557513
N
4333333
Total
1.543803
14541.9
3440932.0
671044191
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